Formulae, Moles, Molarity - Assignment Example

FORMULAE/MOLES/MOLARITY FORMULAE & EQUATIONS Write formulae for the following compounds: a) magnesium chloride - d) sodium hydrogen carbonate -b) calcium carbonate - e) ammonium sulphate - c) potassium phosphate - f) barium nitrate - 2. Write equations for the following reactions, making sure that they are properly balanced. State symbols must be included. a) Potassium hydroxide and nitric acid react together in solution to form potassium nitrate solution and water. b) When heated in a stream of chlorine gas, aluminium metal forms solid, aluminium chloride. c) Lead nitrate and potassium bromide solutions react together to form a precipitate of lead bromide and aqueous potassium nitrate. MOLES 3. Calculate the mass of each of the following: Note: AW means Atomic Weight and the subscript FW stands for Gram Formula Weight. a) 1 mole of NaCl b) 3 moles of CaCO3 c) 0.25 moles of C6H12O6 d) 1 mole of Fe(NO3) 2 e) 1/3 mole of MgSO4.7H2O 4. Calculate the number of moles in each of the following: a) 5.9 g of KBr b) 3.2 g of CO2 c) 33.1 g CuSO4 d) 1.6 kg FeO 5. Calculate the empirical formulae of each of the following compounds: a) the oxide of iron formed when 35 g iron combines with 15 g oxygen. The empirical formula of a substance tells us the relative number of atoms of each element it contains. Because of experimental errors, the results may not lead to exact integers for the ratios of moles. Thus we must multiply 2 to get the integers 2 and 3, for Fe and O subscripts respectively. This is the simplest, or empirical, formula because the subscripts are the smallest integers that express ratios of atoms present in the compound. Empirical Formula: b) the oxide of iron formed when 3.62 g iron reacts with oxygen to form 5.00 g of oxide. The empirical formula of a substance tells us the relative number of atoms of each element it contains. Because of experimental errors, the results may not lead to exact integers for the ratios of moles. Thus we must multiply 3 to get the integers 3 and 4, for Fe and O subscripts respectively. This is the simplest, or empirical, formula because the subscripts are the smallest integers that express ratios of atoms present in the compound. Empirical Formula: 6. When hydrogen chloride gas is passed over sodium hydroxide a reaction occurs, forming sodium chloride and water, as follows: NaOH(s) + HCl(g)  NaCl(s) + H2O(l) The coefficients in a balanced chemical equation can be interpreted as the relative numbers of moles involved in the reaction. a) How many grams of hydrogen chloride are needed to react with 4.0 g NaOH ? b) If excess HCl is passed over 8.0 g NaOH how many grams of NaCl are made? c) How many grams of NaOH must be reacted with excess HCl to make 9.0 g of water? MOLARITY 7. If 0.1 moles of sodium hydroxide are dissolved in l dm3 of water, how many moles will there be in: We are looking for the number of moles in each of the given volumes that maintain the MOLARITY of the solution, which is 0.1M. a) 100 cm3 b) b) 25cm3 8. 25 cm3 of 0.15 mol dm-3 sodium hydroxide reacts with 20cm3 of nitric acid. NaOH + HNO3  NaNO3 + H2O We use the molarity and volume of the solutions to determine the number of moles. Then the coefficients in the balanced equation give the relative number of moles of reactants and products. a) calculate the number of moles of NaOH in the 25 cm3 b) what is the number of moles of nitric acid in 20 cm3 ? c) calculate the molar concentration of the nitric acid 9. 30cm3 of potassium hydroxide reacts with 26.5cm3 of 0.55mol dm-3 sulphuric acid. 2KOH + H2SO4  K2SO4 + 2H2O a) calculate the number of moles of H2SO4 in the 26.5 cm3 b) what is the number of moles of potassium hydroxide in 30 cm3 ? c) calculate the molar concentration of the KOH d) what is the concentration of the potassium hydroxide in g dm-3 ? Electronic Structure & Ionization Energy. 10. Write the electronic structure (in s,p,d notation) of the following: O: Na: Na+: Al: Cl-: Co: 11. Write the electronic configuration (electrons in box notation) of the following: N: Si: Ni: 12. Write the electronic configuration (electrons in box notation) of chromium and copper. Suggest a reason for the apparently anomalous arrangement of electrons in their atoms. Chromium: Copper: Certain elements appear to violate the rules in electron configuration. The electron configuration of chromium is rather than , as one might have expected. Similarly, the configuration of copper is instead of . This anomalous behavior is largely a consequence of the closeness of the 3d and 4s orbital energies. It frequently occurs when there are enough electrons to lead to precisely half-filled sets of degenerate orbitals (as in chromium) or to completely filled d subshell (as in copper). 13. The following table shows the first three ionization energies (in kJ mol-1) of elements in the SAME group of the Periodic Table. Element Hi1 Hi2 Hi3 A 383 2437 3376 B 409 2667 3881 C 425 3065 4438 D 502 4568 6929 E 527 7314 11820 (a) In which group of the Periodic Table should the elements be placed? Give a reason for your answer The elements should be placed under Group IA (Alkali metals) for they have ionization energies within the range of the ionization energies of the alkali group. (b) Which of the elements has the largest atomic number? Give a reason for your answer. Increasing the distance from the nucleus decreases the attraction between the electrons and the nucleus. As this attraction decreases, it becomes easier to remove the electron and, thus, lower ionization energy. Element A has the largest atomic number (greatest distance between outer electron and nucleus) because it has the least ionization energy. 14. The following table shows the first four ionization energies of the elements A to E in kJ mol-1. Element Hi1 Hi2 Hi3 Hi4 A 800 2400 3700 25000 B 740 1500 7700 10500 C 900 1757 14850 21000 D 418 3069 4439 5876 E 577 1816 2745 11575 (a) In which group of the Periodic Table should each element be placed? The ionization energies for an element increase in magnitude as successive electrons are removed. But when the next electron removed is an inner-shell electron, there is a sharp increase in ionization energy. The figure shown below tells us which group elements A-E should be placed. Element A: Group IIIA Element B: Group IIA Element C: Group IIA Element D: Group IA Element E: Group IIIA (b) How much energy is needed to convert one mole of gaseous atoms of element C into 1 mole of dipositive ions? The first ionization energy is the energy needed to remove the first electron from a neutral atom. : I1= 1086.2 kJ/mol The second ionization energy is the energy needed to remove the second electron. : I2 = 2352 kJ/mol ITOTAL = I1 + I2 = 3438.2 kJ/mol MOLARITY CALCULATIONS 15. Calculate number of moles of: a) HCl in 25 cm3 of 0.10 mol dm-3 solution b) H2S04 in 32 cm3 of 0.50 mol dm-3 solution 16. Calculate the volume of a) 0.02 mol dm-3 HCl solution containing 1 X 10-3 moles of HCl b) 0.01 mol dm-3 H2S04 solution containing 2 X 10-3 moles of H2S04 17. Calculate the molar concentrations (in mol dm-3) of the following solutions that contain: a) 2.5 x 10-3 moles AgNO3 in 25 cm3 b) 5 x 10-3 moles BaCI2 in 10 cm3 c 18. 23 cm3 of 1.5 mol dm-3 H2SO4 reacts completely with 42.5 cm3 of a given KOH solution. 2KOH + H2SO4  K2SO4 + 2H2O What is the molar concentration of the KOH solution? 19. 27.823g of Na2CO3.xH2O crystals were dissolved in water and made up to 1000 cm3 of solution. 25 cm3 of this solution required 48.8 cm3 of 0.1 mol dm-3 HCI for complete neutralisation. Find the value of x in Na2CO3.xH2O using the following steps: 2HCl + Na2CO3  2NaCl + CO2 + H2O a) calculate the number of moles of HCl in the 48.8 cm3 b) calculate the number of moles of Na2CO3 in 25 cm3 c) calculate the number of moles of Na2CO3 in 1000 cm3 The molarity of the 1000 cm3 solution is the same as that of the 25 cm3 solution since the 25 cm3 solution was taken from the 1000 cm3 solution. d) calculate the mass of Na2CO3 in 1000 cm3 e) calculate the mass of water of crystallization associated with this mass of Na2CO3 f) calculate the moles of water of crystallization associated with this mass of Na2CO3 g) calculate the value of x in Na2CO3.xH2O There are ten moles of water for every mole of sodium carbonate: 20. 4.00g of lawn sand (a mixture of sand and ammonium sulphate) was weighed into a conical flask, and 25 cm3 of 2.0 mol dm-3 sodium hydroxide solution was pipetted into the same flask. The conical flask was boiled for 20 minutes, after which time all the ammonia had been driven off, because: (NH4) 2SO4 (s) + 2NaOH (aq)  2NH3 (g) + Na2SO4(aq) + 2H2O (1) The residue in the flask was cooled and filtered to remove the sand. The filtrate containing unreacted NaOH was made up to 250 cm3 in a volumetric flask. 25 cm3 samples of this solution were titrated against 0.1 mol dm-3 hydrochloric acid using bromothymol blue as an indicator. HCl + NaOH  NaCl + H2O The mean titre was 20.0 cm3. Calculate the percentage of ammonium sulphate by mass in the lawn sand. From HCl + NaOH  NaCl + H2O, From (NH4) 2SO4 (s) + 2NaOH (aq)  2NH3 (g) + Na2SO4(aq) + 2H2O (1), Source: Brown, T., Bursten, B., Lemay, E. (2002). Chemistry: The central science. New York: Prentice Hall Read More